Đáp án:

$1,9872(g)$

Giải thích các bước giải:

$V_{C_2H_5OH}=\frac{4,6.46}{100}=2,116(ml)$

$\to m_{C_2H_5OH}=2,116.0,8=1,6928(g)$

$\to n_{C_2H_5OH}=\frac{1,6928}{46}=0,0368(mol)$

$\to n_{C_2H_5OH(pứ)}=0,0368.90\%=0,03312(mol)$

$C_2H_5OH+O_2\xrightarrow{men}CH_3COOH+H_2O$

Theo PT: $n_{CH_3COOH}=n_{C_2H_5OH(pứ)}=0,03312(mol)$

$\to m_{CH_3COOH}=0,03312.60=1,9872(g)$

Đáp án:

`m_{CH_3COOH}=1,9872` `gam`

Giải thích các bước giải:

`V_{C_2H_5OH}=``\frac{4,6.46}{100}=2,116` `(l)`

`m_{C_2H_5OH}=D.V=0,8.2,116=1,6928` `(g)`

`n_{C_2H_5OH}=``\frac{m}{M}=\frac{1,6928}{46}=0,0368` `(mol)`

`PTHH`         `C_2H_5OH``+``O_2` $\xrightarrow[]{men giấm}$ `CH_3COOH``+``H_2O`

                      `0,0368`                          `0,0368`                       `mol`

`→n_{CH_3COOH}=n_{C_2H_5OH}=0,0368` `(mol)`

Vì `H=90%`

Nên `m_{CH_3COOH}=0,0368.60.\frac{90}{100}=1,9872` `(g)`